On Chordal-$k$-Generalized Split Graphs

TL;DR

This paper characterizes chordal-$k$-generalized split graphs through forbidden subgraphs and explores the computational complexity of the Efficient Domination problem in a special case.

Contribution

It provides a forbidden subgraph characterization for chordal-$k$-generalized split graphs and analyzes the NP-completeness of the Efficient Domination problem in a specific case.

Findings

Chordal-$k$-generalized split graphs are characterized by forbidden induced subgraphs.

The paper identifies a special case of chordal-$2$-generalized split graphs where the Efficient Domination problem is NP-complete.

Split graphs are exactly the chordal-$1$-generalized split graphs.

Abstract

A graph $G$ is a {\em chordal-$k$-generalized split graph} if $G$ is chordal and there is a clique $Q$ in $G$ such that every connected component in $G[V \setminus Q]$ has at most $k$ vertices. Thus, chordal-$1$-generalized split graphs are exactly the split graphs. We characterize chordal-$k$-generalized split graphs by forbidden induced subgraphs. Moreover, we characterize a very special case of chordal-$2$-generalized split graphs for which the Efficient Domination problem is \NP-complete.

Full text

We characterize chordal-$k$-generalized split graphs by forbidden induced subgraphs. Moreover, we characterize a very special case of chordal-$2$-generalized split graphs for which the Efficient Domination problem is $\mathbb{NP}$-complete.

**Keywords: Chordal graphs; split graphs; partitions into a clique and connected components of bounded size; forbidden induced subgraphs; $\mathbb{NP}$-completeness; polynomial time recognition. **

1 Introduction

$G=(V,E)$ is a split graph if $V$ can be partitioned into a clique and an independent set. The famous class of split graphs was characterized by Főldes and Hammer in [5] as the class of $2K_{2}$-free chordal graphs, i.e., the class of $(2K_{2},C_{4},C_{5})$-free graphs, and $G$ is a split graph if and only if $G$ and its complement graph $\overline{G}$ are chordal.

There are various important kinds of generalizations of split graphs: $G$ is unipolar if there is a clique $Q$ in $G$ such that $G[V\setminus Q]$ is the disjoint union of cliques of $G$, i.e., $G[V\setminus Q]$ is $P_{3}$-free. Clearly, not every unipolar graph is chordal. $G$ is called a generalized split graph [8] if either $G$ or $\overline{G}$ is unipolar. Generalized split graphs are perfect, and Prömel and Steger [8] showed that almost all perfect graphs are generalized split graphs.

We consider a different kind of generalizing split graphs: A graph $G$ is a chordal-$k$-generalized split graph if $G$ is chordal and there is a clique $Q$ in $G$ such that every connected component in $G[V\setminus Q]$ has at most $k$ vertices; we call such a clique $Q$ a $k$-good clique of $G$. Thus, chordal-$1$-generalized split graphs are exactly the split graphs. We characterize chordal-$k$-generalized split graphs by forbidden induced subgraphs.

An induced matching $M\subseteq E$ is a set of edges whose pairwise distance in $G$ is at least two. A hereditary induced matching (h.i.m.) is the induced subgraph of an induced matching, i.e., every connected component has at most two vertices, i.e., it is the disjoint union of an independent vertex set $S$ and the vertex set of an induced matching $M$ in $G$.

Thus, $G$ is a chordal-$2$-generalized split graph if and only if $G$ has a clique $Q$ such that $G[V\setminus Q]$ is a hereditary induced matching.

2 Basic Notions and Results

For a vertex $v\in V$, $N(v)=\{u\in V:uv\in E\}$ denotes its (open) neighborhood, and $N[v]=\{v\}\cup N(v)$ denotes its closed neighborhood. A vertex $v$ sees the vertices in $N(v)$ and misses all the others. The non-neighborhood of a vertex $v$ is $\overline{N}(v):=V\setminus N[v]$. For $U\subseteq V$, $N(U):=\bigcup_{u\in U}N(u)\setminus U$ and $\overline{N}(U):=V\setminus(U\cup N(U))$.

For a set ${\cal F}$ of graphs, a graph $G$ is called ${\cal F}$-free if $G$ contains no induced subgraph isomorphic to a member of ${\cal F}$. In particular, we say that $G$ is $H$-free if $G$ is $\{H\}$-free. Let $H_{1}+H_{2}$ denote the disjoint union of graphs $H_{1}$ and $H_{2}$, and for $k\geq 2$, let $kH$ denote the disjoint union of $k$ copies of $H$. For $i\geq 1$, let $P_{i}$ denote the chordless path with $i$ vertices, and let $K_{i}$ denote the complete graph with $i$ vertices (clearly, $P_{2}=K_{2}$). For $i\geq 4$, let $C_{i}$ denote the chordless cycle with $i$ vertices. A graph $G$ is chordal if it is $C_{i}$-free for any $i\geq 4$.

$H$ is a butterfly (see Figure 1) if $H$ has five vertices, say $v_{1},\ldots,v_{5}$ such that $v_{1},v_{2},v_{3}$ induce a $K_{3}$ and $v_{3},v_{4},v_{5}$ induce a $K_{3}$, and there is no other edge in $H$. $H$ is an extended butterfly (see Figure 1) if $H$ has six vertices, say $v_{1},\ldots,v_{6}$ such that $v_{1},v_{2},v_{3}$ induce a $K_{3}$ and $v_{4},v_{5},v_{6}$ induce a $K_{3}$, and the only other edge in $H$ is $v_{3}v_{4}$.

$H$ is an extended co-$P$ (see Figure 1) if $H$ has six vertices, say $v_{1},\ldots,v_{6}$ such that $v_{1},\ldots,v_{5}$ induce a $P_{5}$ and $v_{6}$ is only adjacent to $v_{1}$ and $v_{2}$.

$H$ is a chair if $H$ has five vertices, say $v_{1},\ldots,v_{5}$ such that $v_{1},\ldots,v_{4}$ induce a $P_{4}$ and $v_{5}$ is only adjacent to $v_{3}$. $H$ is an extended chair (see Figure 1) if $H$ has six vertices, say $v_{1},\ldots,v_{6}$ such that $v_{1},\ldots,v_{5}$ induce a chair as above and $v_{6}$ is only adjacent to $v_{1}$ and $v_{2}$.

$H$ is a gem if $H$ has five vertices, say $v_{1},\ldots,v_{5}$ such that $v_{1},\ldots,v_{4}$ induce a $P_{4}$ and $v_{5}$ is adjacent to $v_{1},v_{2},v_{3}$, and $v_{4}$.

$H$ is a double-gem (see Figure 1) if $H$ has six vertices, say $v_{1},\ldots,v_{6}$ such that $v_{1},\ldots,v_{5}$ induce a gem, say with $P_{4}$ $(v_{1},v_{2},v_{3},v_{4})$ and $v_{6}$ is only adjacent to $v_{3}$ and $v_{4}$.

We say that for a vertex set $X\subseteq V$, a vertex $v\notin X$ has a join (resp., co-join) to $X$ if $X\subseteq N(v)$ (resp., $X\subseteq\overline{N}(v)$). Join (resp., co-join) of $v$ to $X$ is denoted by $v\text{{\footnotesize1}⃝}X$ (resp., $v\text{{\footnotesize0}⃝}X$); if $v\text{{\footnotesize1}⃝}X$ then $v$ is universal for $X$. Correspondingly, for vertex sets $X,Y\subseteq V$ with $X\cap Y=\emptyset$, $X\text{{\footnotesize1}⃝}Y$ denotes $x\text{{\footnotesize1}⃝}Y$ for all $x\in X$ and $X\text{{\footnotesize0}⃝}Y$ denotes $x\text{{\footnotesize0}⃝}Y$ for all $x\in X$. A vertex $x\notin U$ contacts $U$ if $x$ has a neighbor in $U$. For vertex sets $U,U^{\prime}$ with $U\cap U^{\prime}=\emptyset$, $U$ contacts $U^{\prime}$ if there is a vertex in $U$ contacting $U^{\prime}$.

As a first step for our results, we show:

Lemma 1**.**

If $G=(V,E)$ is a chordal graph, $Q$ is a clique in $G$ and $Z$ is a connected component in $G[V\setminus Q]$ such that every vertex $q\in Q$ has a neighbor in $Z$ then there is a universal vertex $z\in Z$ for $Q$, i.e., $z\text{{\footnotesize1}⃝}Q$.

Proof. We show inductively for $Q=\{q_{1},\ldots,q_{k}\}$ that there is a $z\in Z$ with $z\text{{\footnotesize1}⃝}Q$:

If $k=1$ then trivially, there is such a vertex $z\in Z$, and if $k=2$, say $Q=\{q_{1},q_{2}\}$, then let $x,y\in Z$ be such that $q_{1}x\in E$ and $q_{2}y\in E$. If $xq_{2}\notin E$ and $yq_{1}\notin E$ then take a shortest path $P_{xy}$ between $x$ and $y$ in $Z$; clearly, since $G$ is chordal, there is a vertex in $P_{xy}$ seeing both $q_{1}$ and $q_{2}$.

Now assume that the claim is true for clique $Q\setminus\{q_{k}\}$ with $k-1$ vertices and let $x\in Z$ be a universal vertex for $Q\setminus\{q_{k}\}$. If $xq_{k}\in E$, we are done. Thus assume that $xq_{k}\notin E$, and let $y\in Z$ with $q_{k}y\in E$. If $xy\in E$ then, since $G$ is chordal, clearly, $y$ is adjacent to all $q_{i}$, $1\leq i\leq k-1$. Thus, assume that the distance between $x$ and $y$ in $Z$ is larger than 1. Without loss of generality, let $y$ be a neighbor of $q_{k}$ in $Z$ which is closest to $x$. Let $P_{xy}$ be a shortest path between $x$ and $y$ in $Z$. Thus, $q_{k}$ is nonadjacent to every internal vertex in $P_{xy}$ and to $x$. Then, since $G$ is chordal, $y$ is adjacent to all $q_{i}$, $1\leq i\leq k-1$. Thus, Lemma 1 is shown. ∎

3 $k$-Good Cliques in Chordal Graphs

3.1 $2K_{2}$-free chordal graphs

For making this manuscript self-contained, let us repeat the characterization of split graphs (and a proof variant which can be generalized for $k$-good cliques).

Theorem 1** ([5]).**

A chordal graph $G$ is $2K_{2}$-free if and only if $G$ is a split graph.

Proof. If $G$ is a split graph then clearly, $G$ is $2K_{2}$-free chordal.

For the converse direction, assume to the contrary that for every clique $Q$ of $G$, there is a connected component, say $Z_{Q}$, of $G[V\setminus Q]$ with at least two vertices; we call such components $2$-nontrivial. Since $G$ is $2K_{2}$-free, all other components of $G[V\setminus Q]$ consist of isolated vertices.

Let $Q_{1}:=\{q\in Q:q$ has a neighbor in $Z_{Q}\}$, and $Q_{2}:=Q\setminus Q_{1}$, i.e., $Q_{2}\text{{\footnotesize0}⃝}Z_{Q}$. Thus, $Q=Q_{1}\cup Q_{2}$ is a partition of $Q$. Since $G$ is $2K_{2}$-free and $Z_{Q}$ is 2-nontrivial, clearly, $|Q_{2}|\leq 1$, and there is no connected component in $G[V\setminus(Q_{1}\cup Z_{Q})]$ with at least two vertices. Thus, $G[V\setminus(Q_{1}\cup Z_{Q})]$ is an independent vertex set.

Let $Q$ be a clique in $G$ with smallest 2-nontrivial component $Z_{Q}$ of $G[V\setminus Q]$ with respect to all cliques in $G$. Clearly, $Z_{Q}$ is also the nontrivial component of $G[V\setminus Q_{1}]$, i.e., $Z_{Q_{1}}=Z_{Q}$. Thus, $Q_{1}$ is a clique in $G$ with smallest 2-nontrivial component, and from now on, we can assume that every vertex in $Q_{1}$ has a neighbor in $Z_{Q_{1}}$. Thus, by Lemma 1, there is a universal vertex $z\in Z_{Q_{1}}$ for $Q_{1}$, i.e., $z\text{{\footnotesize1}⃝}Q_{1}$.

This implies that for the clique $Q^{\prime}:=Q_{1}\cup\{z\}$, the 2-nontrivial component $Z_{Q^{\prime}}$ (if there is any for $Q^{\prime}$) is smaller than the one of $Q_{1}$ which is a contradiction. Thus, Theorem 1 is shown. ∎

3.2 $(2P_{3},2K_{3},P_{3}+K_{3})$-free chordal graphs

Clearly, a connected component with three vertices is either a $P_{3}$ or $K_{3}$, and graph $G$ is $(2P_{3},2K_{3},P_{3}+K_{3})$-free chordal if and only if $G$ is $(2P_{3},2K_{3},P_{3}+K_{3},C_{4},C_{5},C_{6},C_{7})$-free. In a very similar way as for Theorem 1, we show:

Theorem 2**.**

A chordal graph $G$ is $(2P_{3},2K_{3},P_{3}+K_{3})$-free if and only if $G$ is a chordal-$2$-generalized split graph.

Proof. If $G$ is a chordal-$2$-generalized split graph then clearly, $G$ is $(2P_{3},2K_{3},P_{3}+K_{3})$-free chordal.

For the converse direction, assume to the contrary that for every clique $Q$ of $G$, there is a connected component, say $Z_{Q}$, of $G[V\setminus Q]$ with at least three vertices; we call such components $3$-nontrivial.

Let $Q_{1}:=\{q\in Q:q$ has a neighbor in $Z_{Q}\}$, and $Q_{2}:=Q\setminus Q_{1}$, i.e., $Q_{2}\text{{\footnotesize0}⃝}Z_{Q}$. Thus, $Q=Q_{1}\cup Q_{2}$ is a partition of $Q$. Since $G$ is $(2P_{3},2K_{3},P_{3}+K_{3})$-free and $Z_{Q}$ is 3-nontrivial, clearly, $|Q_{2}|\leq 2$, and there is no connected component in $G[V\setminus(Q_{1}\cup Z_{Q})]$ with at least three vertices. Thus, $G[V\setminus(Q_{1}\cup Z_{Q})]$ is a hereditary induced matching.

Let $Q$ be a clique in $G$ with smallest 3-nontrivial component $Z_{Q}$ of $G[V\setminus Q]$ with respect to all cliques in $G$. Clearly, $Z_{Q}$ is also the nontrivial component of $G[V\setminus Q_{1}]$, i.e., $Z_{Q_{1}}=Z_{Q}$. Thus, $Q_{1}$ is a clique in $G$ with smallest 3-nontrivial component, and from now on, we can assume that every vertex in $Q_{1}$ has a neighbor in $Z_{Q_{1}}$. Thus, by Lemma 1, there is a universal vertex $z\in Z_{Q_{1}}$ for $Q_{1}$, i.e., $z\text{{\footnotesize1}⃝}Q_{1}$. This implies that for the clique $Q^{\prime}:=Q_{1}\cup\{z\}$, the 3-nontrivial component $Z_{Q^{\prime}}$ (if there is any for $Q^{\prime}$) is smaller than the one of $Q_{1}$ which is a contradiction. Thus, Theorem 2 is shown. ∎

Clearly, not every $(2P_{3},2K_{3},P_{3}+K_{3})$-free graph and even not every $(2P_{3}$, $2K_{3}$, $P_{3}+K_{3}$, $C_{5},C_{6},C_{7})$-free graph has a 3-good clique as the following example shows:

Let $v_{1},v_{2},v_{3},v_{4}$ induce a $C_{4}$ with edges $v_{i}v_{i+1}$ (index arithmetic modulo 4) and let $x_{1},x_{2},x_{3}$ be personal neighbors of $v_{1},v_{2},v_{3}$ correspondingly. Then any clique $Q$ has only at most two vertices, and for none of them, $G[V\setminus Q]$ is a hereditary induced matching.

3.3 The general case of $h$-good cliques in chordal graphs

As usual, for a pair of connected graphs $H_{1}$, $H_{2}$ with disjoint vertex sets, let $H_{1}+H_{2}$ denote the disjoint union of $H_{1}$ and $H_{2}$. For any natural $h$, let ${\cal C}_{h}$ denote the family of connected graphs with $h$ vertices, and let ${\cal A}_{h}=\{X+Y:X,Y\in{\cal C}_{h}\}$. In a very similar way as for Theorems 1 and 2, we can show:

Theorem 3**.**

For any natural $h$, chordal graph $G$ is ${\cal A}_{h+1}$-free if and only if there is a clique $Q$ of $G$ such that every connected component of $G[V\setminus Q]$ has at most $h$ vertices.

Proof. If $G$ is a chordal graph with a clique $Q$ of $G$ such that every connected component of $G[V\setminus Q]$ has at most $h$ vertices, then clearly, $G$ is ${\cal A}_{h+1}$-free chordal.

For the converse direction, assume to the contrary that for every clique $Q$ of $G$, there is a connected component, say $Z_{Q}$, of $G[V\setminus Q]$ with at least $h+1$ vertices; we call such components $(h+1)$-nontrivial.

Let $Q_{1}:=\{q\in Q:q$ has a neighbor in $Z_{Q}\}$, and $Q_{2}:=Q\setminus Q_{1}$, i.e., $Q_{2}\text{{\footnotesize0}⃝}Z_{Q}$. Thus, $Q=Q_{1}\cup Q_{2}$ is a partition of $Q$. Since $G$ is ${\cal A}_{h+1}$-free and $Z_{Q}$ is $(h+1)$-nontrivial, clearly, $|Q_{2}|\leq h$, and there is no connected component in $G[V\setminus(Q_{1}\cup Z_{Q})]$ with at least $h+1$ vertices. Thus, every connected component of $G[V\setminus(Q_{1}\cup Z_{Q})]$ has at most $h$ vertices.

Let $Q$ be a clique in $G$ with smallest $(h+1)$-nontrivial component $Z_{Q}$ of $G[V\setminus Q]$ with respect to all cliques in $G$. Clearly, $Z_{Q}$ is also the nontrivial component of $G[V\setminus Q_{1}]$, i.e., $Z_{Q_{1}}=Z_{Q}$. Thus, $Q_{1}$ is a clique in $G$ with smallest $(h+1)$-nontrivial component, and from now on, we can assume that every vertex in $Q_{1}$ has a neighbor in $Z_{Q_{1}}$. Thus, by Lemma 1, there is a universal vertex $z\in Z_{Q_{1}}$ for $Q_{1}$, i.e., $z\text{{\footnotesize1}⃝}Q_{1}$.

This implies that for the clique $Q^{\prime}:=Q_{1}\cup\{z\}$, the $(h+1)$-nontrivial component $Z_{Q^{\prime}}$ (if there is any for $Q^{\prime}$) is smaller than the one of $Q_{1}$ which is a contradiction. Thus, Theorem 3 is shown. ∎

Concerning the recognition problem, Tyshkevich and Chernyak [9] showed that unipolar graphs can be recognized in ${\cal O}(n^{3})$ time. This time bound was slightly improved in [4], and in [7], McDiarmid and Yolov give a ${\cal O}(n^{2})$ time recognition algorithm for unipolar graphs and generalized split graphs. Clearly, for each fixed $k$, chordal-$k$-generalized split graphs can be recognized in ${\cal O}(nm)$ time since chordal graphs have at most $n$ maximal cliques, and for each of them, say $Q$, it can be checked in linear time whether the connected components of $G[V\setminus Q]$ have at most $k$ vertices.

4 Characterizing Split-Matching-Extended Graphs

$G=(V,E)$ is a split-matching-extended graph if $V$ can be partitioned into a clique $Q$, an independent vertex set $S_{Q}$ and the vertex set of an induced matching $M_{Q}$ in $G$ such that $S_{Q}\text{{\footnotesize0}⃝}V(M_{Q})$ and for every edge $xy\in M_{Q}$, at most one of $x,y$ has neighbors in $Q$ (note that $S_{Q}\cup V(M_{Q})$ is a hereditary induced matching in $G$). Clearly, split-matching-extended graphs are chordal. Thus, split-matching-extended graphs are a special case of chordal-2-generalized split graphs.

Clearly, split-matching-extended graphs can be recognized in linear time since for every edge $xy\in M$, the degree of $x$ or $y$ is 1, and deleting all such vertices of degree 1 leads to a split graph (which can be recognized in linear time).

Various algorithmic problems such as Hamiltonian Circuit and Minimum Domination are $\mathbb{NP}$-complete for split graphs. Efficient Domination, however, is solvable in polynomial time (even in linear time [1]) for split graphs while it is $\mathbb{NP}$-complete for split-matching-extended graphs (see e.g. [2]).

As mentioned in [2], the graph $G_{H}$ in the reduction of Exact Cover 3XC to Efficient Domination is $(2P_{3}$, $K_{3}+P_{3}$, $2K_{3}$, butterfly, extended butterfly, extended co-$P$, extended chair, double-gem$)$-free chordal (and clearly, a special kind of unipolar graphs). Clearly, $2P_{3}$-free implies $C_{k}$-free for each $k\geq 8$.

Theorem 4**.**

$G$* is a split-matching-extended graph if and only if $G$ is $(C_{4}$, $C_{5}$, $C_{6}$, $C_{7}$, $2P_{3}$, $K_{3}+P_{3}$, $2K_{3}$, butterfly, extended butterfly, extended co-$P$, extended chair, double-gem$)$-free.*

Proof. Clearly, if $G$ is a split-matching-extended graph then $G$ is $(C_{4}$, $C_{5}$, $C_{6}$, $C_{7}$, $2P_{3}$, $K_{3}+P_{3}$, $2K_{3}$, butterfly, extended butterfly, extended co-$P$, extended chair, double-gem$)$-free.

For the other direction, assume that $G$ is $(C_{4},C_{5},C_{6},C_{7},2P_{3}$, $K_{3}+P_{3}$, $2K_{3}$, butterfly, extended butterfly, extended co-$P$, extended chair, double-gem$)$-free. Then by Theorem 2, $G$ has a 2-good clique $Q$ such that $G[V\setminus Q]$ is a hereditary induced matching, say with independent vertex set $S_{Q}=\{s_{1},\ldots,s_{k}\}$ and induced matching $M_{Q}=\{x_{1}y_{1},\ldots,x_{\ell}y_{\ell}\}$. Let $Q$ be a maximum 2-good clique of $G$.

Claim 4.1**.**

For every edge $xy\in E$ with $x,y\in N(Q)$, $N_{Q}(x)\subseteq N_{Q}(y)$ and correspondingly for the non-neighborhoods, $\overline{N_{Q}}(y)\subseteq\overline{N_{Q}}(x)$ or vice versa, and in particular, $x$ and $y$ have a common neighbor and a common non-neighbor in $Q$.

Proof. The claim obviously holds since $G$ is chordal and $Q$ is maximum. $\diamond$

Claim 4.2**.**

For every edge $xy\in E$ with $x,y\in N(Q)$, $x$ and $y$ have exactly one common non-neighbor in $Q$.

Proof. Recall that by Claim 4.1, $x$ and $y$ have a common neighbor, say $q_{xy}\in Q$, and suppose to the contrary that $x$ and $y$ have two common non-neighbors, say $q_{1},q_{2}\in Q$, $q_{1}\neq q_{2}$. Then $x,y,q_{xy},q_{1},q_{2}$ induce a butterfly in $G$ which is a contradiction. Thus, Claim 4.2 is shown. $\diamond$

Claim 4.3**.**

For every edge $xy\in E$ with $x,y\in N(Q)$, at least one of $x$ and $y$ has at least two non-neighbors in $Q$. Moreover, if $N_{Q}(x)\subseteq N_{Q}(y)$ then $y$ has exactly one non-neighbor in $Q$, and thus, $N_{Q}(x)\subset N_{Q}(y)$.

Proof. If both $x$ and $y$ would have exactly one non-neighbor, say $q_{x},q_{y}\in Q$ with $xq_{x}\notin E$ and $yq_{y}\notin E$ then by Claim 4.1, $q_{x}=q_{y}$ and now, $(Q\setminus\{q_{x}\})\cup\{x,y\}$ would be a larger clique in $G$ but $Q$ is assumed to be a maximum clique in $G$. Thus, at least one of $x,y$ has at least two non-neighbors in $Q$.

By Claim 4.1, we have the corresponding inclusions of neighborhoods and non-neighborhoods in $Q$. If $N_{Q}(x)\subseteq N_{Q}(y)$ and $y$ has two non-neighbors in $Q$ then also $x$ has these non-neighbors but this contradicts Claim 4.2. $\diamond$

Claim 4.4**.**

$G[N(Q)]$* is $2K_{2}$-free.*

Proof. Suppose to the contrary that $G[N(Q)]$ contains $2K_{2}$, say $xy,x^{\prime}y^{\prime}$ with $xy\in E$ and $x^{\prime}y^{\prime}\in E$. Let $z\in Q$ be a common neighbor of $x,y$, and let $z^{\prime}\in Q$ be a common neighbor of $x^{\prime},y^{\prime}$. If $z=z^{\prime}$ then $x,y,x^{\prime},y^{\prime},z$ induce a butterfly. Thus, $z\neq z^{\prime}$ for any common neighbors $z$ of $xy$, $z^{\prime}$ of $x^{\prime}y^{\prime}$. If $x^{\prime},y^{\prime}$ miss $z$ and $x,y$ miss $z^{\prime}$ then $x,y,x^{\prime},y^{\prime},z,z^{\prime}$ induce an extended butterfly. If $x^{\prime},y^{\prime}$ miss $z$ but exactly one of $x,y$, say $x$, sees $z^{\prime}$ then $x,z,x^{\prime},y^{\prime},z^{\prime}$ induce a butterfly. If exactly one of $x,y$, say $x$, sees $z^{\prime}$ and exactly one of $x^{\prime},y^{\prime}$, say $x^{\prime}$, sees $z$ then $x,y,z,z^{\prime},x^{\prime},y^{\prime}$ induce a double-gem. Thus, in any case we get a contradiction, and Claim 4.4 is shown. $\diamond$

Thus, for at most one of the edges $x_{i}y_{i}\in M_{Q}$, both $x_{i}$ and $y_{i}$ have a neighbor in $Q$; without loss of generality, assume that $x_{1}$ and $y_{1}$ have a neighbor in $Q$, and for all other edges $x_{j}y_{j}\in M_{Q}$, $j\geq 2$, $y_{j}\text{{\footnotesize0}⃝}Q$. By Claim 4.1, let without loss of generality $N_{Q}(x_{1})\subseteq N_{Q}(y_{1})$, and let $u\in Q$ be a common neighbor of $x_{1}$ and $y_{1}$. By Claim 4.3, $y_{1}$ has exactly one non-neighbor, say $z$, in $Q$ (which also misses $x_{1}$). Since $G$ is extended-co-$P$-free and butterfly-free, and by definition, $z\text{{\footnotesize0}⃝}\{y_{1},\ldots,y_{\ell}\}$, we have $z\text{{\footnotesize0}⃝}\{x_{1},\ldots,x_{\ell}\}$, and since $G$ is extended-chair-free and butterfly-free, $z$ has at most one neighbor in $S_{Q}$.

Let $Q^{\prime}:=(Q\setminus\{z\})\cup\{y_{1}\}$; clearly, $Q^{\prime}$ is a clique in $G$. Then according to $Q^{\prime}$, $G$ is a split-matching-extended graph, and thus, Theorem 4 is shown. ∎

Note added in proof. After writing this manuscript, we learnt that actually, the results of this manuscript are not new; they follow from previous papers of Gagarin [6] and of Zverovich [10]. Sorry for that!